Find last digit.(1023^3923)+(3087^3927) (what is d tricks of finding these last digit sums??..plzz explain)

• last digit:
  Cycle of 3:
  1) 3^1= 3 is the last digit.
  2) 3^2= 9 is the last digit.
  3) 3^3= 7 is the last digit.
  4) 3^4= 1 is the last digit.
  an then for 3^5 onward's the cycle repeats.
  Therefore, for 1023^3923 => 3^3923 = 3^(4*980 +3) So remainder is 3 when the exponent 3923 is divided by 4. So from point '4)' we move 3 steps in cyclic order to go to point '3)'. So last digit of the first part is 7.
  Similarly cycle for 7 for the 2nd part.
  1) 7^1= 7 is the last digit.
  2) 7^2= 9 is the last digit.
  3) 7^3= 3 is the last digit.
  4) 7^4= 1 is the last digit.
  an then for 7^5 onward's the cycle repeats.
  And hence find the last digit in the similar fashion. It comes to 3.
  So total is 7+3= 10. So last digit is 0.
• 10 years agoHelpfull: Yes(29) No(0)
• 1023^3923(3923=4n+3) so last digit 7
  3087^3927(3927=4n+3)so last digit 3
  now adding we get last digit 0
• 10 years agoHelpfull: Yes(7) No(1)
• last digit:
  Cycle of 3:
  1) 3^1= 3 is the last digit.
  2) 3^2= 9 is the last digit.
  3) 3^3= 7 is the last digit.
  4) 3^4= 1 is the last digit.
  an then for 3^5 onward's the cycle repeats.
  so cyclic for 3 is 4.
  Therefore, for 1023^3923 => 3^3923 = 3^(4*980 +3),3^3=27 So last digit of the first part is 7.
  Similarly cycle for 7 for the 2nd part.
  1) 7^1= 7 is the last digit.
  2) 7^2= 9 is the last digit.
  3) 7^3= 3 is the last digit.
  4) 7^4= 1 is the last digit.
  so cyclic of 7 is 4.
  And hence find the last digit in the similar fashion. It comes to 3.
  So total is 7+3= 10. So last digit is 0.
• 10 years agoHelpfull: Yes(3) No(0)
• 1st part last digit is 7 as 3^3. 2nd part: 7^3 as(4 is universil cyclicty. so 7 didided by 4= 3. and 7^3=lst digit 3.). so 7+3=0 so ans 0 is d last digit.
• 10 years agoHelpfull: Yes(1) No(0)
• last digit is 0
• 10 years agoHelpfull: Yes(1) No(0)
• lst digit of frst term=7, lst digit of 2nd term 3, 7+3=10...overall lst digit=0
• 10 years agoHelpfull: Yes(1) No(0)
• LAST DIGIT:
  Cycle of 3:
  1) 3^1= 3 is the last digit.
  2) 3^2= 9 is the last digit.
  3) 3^3= 7 is the last digit.
  4) 3^4= 1 is the last digit.
  an then for 3^5 onward's the cycle repeats.
  Therefore, for 1023^3923 => 3^3923 = 3^(4*980 +3) So remainder is 3 when the exponent 3923 is divided by 4. So from point '4)' we move 3 steps in cyclic order to go to point '3)'. So last digit of the first part is 7.
  Similarly cycle for 7 for the 2nd part.
  1) 7^1= 7 is the last digit.
  2) 7^2= 9 is the last digit.
  3) 7^3= 3 is the last digit.
  4) 7^4= 1 is the last digit.
  an then for 7^5 onward's the cycle repeats.
  And hence find the last digit in the similar fashion. It comes to 3.
  So total is 7+3= 10. So last digit is 0.
• 10 years agoHelpfull: Yes(1) No(0)
• ans is 4. the last digits when 3 is raised to the following powers:
  3^1=3
  3^2=9
  3^3=7
  3^4=1
  3^5=3
  3^6=9
  3^7=7
  3^8=1. thus cyclicity of 3 is 4. Dividing the power by 4 we get ((1023^4)^980)*1023^3. Thus last digit in this case is 7. Similarly for the second part the last digit obtained s 7. Hence the overall last digit is 4
  
• 10 years agoHelpfull: Yes(0) No(10)
• 4
  1023^((5*784)+3)=......1
  3087^((5*785)+2)=......3
  1+3=4
• 10 years agoHelpfull: Yes(0) No(1)
• 23+87=110 so last digits=0
• 10 years agoHelpfull: Yes(0) No(1)
• laast digit is "0"
• 10 years agoHelpfull: Yes(0) No(1)